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Chemistry Question on Galvanic Cells

In the cell Pt(s)∣H2(g,1bar∣HCl(aq)∣Ag(s)∣Pt(s){Pt(s) | H_2(g , 1 bar | HCl(aq) | Ag(s) | Pt(s)} the cell potential is 0.92 when a 10−610^{-6} molal HCl solution is used. THe standard electrode potential of (AgCl/Ag,Cl−){(AgCl/Ag,Cl^{-})} electrode is : \left\\{ \text{given} , \frac{2.303RT}{F} = 0.06 V \; at \; 298K\right\\}

A

0.20 V

B

0.76 V

C

0.40 V

D

0.94 V

Answer

0.20 V

Explanation

Solution

Pt(s)H2(g,1bar)HCl(aq)AgCl(s)Ag(s)∣Pt(s){Pt(s) H_2 (g, 1bar) HCl(aq) AgCl(s) Ag(s) |Pt(s) }
Anode: H2−>[10−6m]2H++2e×1{H_2 ->[10^{-6} m] 2H^{+} + 2e \times 1}
Cathode : e−+AgCl(s)−>Ag(s)+Cl−(aq){e^{-} + AgCl(s) -> Ag(s) + Cl^{-} (aq) }
H2(g)l+AgCl(s)−>2H++2Ag(s)+2Cl−(aq){H_2 (g) l + AgCl(s) -> 2H^{+} + 2Ag(s) + 2 Cl^{-} (aq)}
Ecell=Ecell0−0.062log⁡10((H+)2.(Cl−)2){E_{cell} = E^{0}_{cell} - \frac{0.06}{2} \log_{10} ((H^{+})^{2} . (Cl^{-})^{2})}
{.925 = (E^{0}_{H_2 /H^{+}}} + E^{0}_{AgCl / Ag, Cl^{-}}) - \frac{0.06}{2} \log_{10} ((10^{-6})^{2} (10^{-6})^{2})}
.92=0+EAgCl/Ag,Cl−0−0.03log⁡10(10−6)4.92 = 0 + E^{0}_{AgCl /Ag, Cl^{-}} - 0.03 \log_{10} (10^{-6} )^4
EAgCl0/Ag,Cl−=.9.2+.03×−24=0.2  V{E^{0}_{AgCl } / Ag, Cl^{-} = .9.2 + .03 \times -24 = 0.2 \; V }